BOOK V. REGULAR POLYGONS. — MEASUREMENT OF THE CIRCLE. 339. Def. A regular polygon is a polygon which is both equilateral and equiangular. 340. A circle can be circumscribed about, or inscribed in, any regular polygon. Given regular polygon ABCDE. To Prove that a O can be circumscribed about, or inscribed in, ABCDE. Proof. Let O be the centre of the circumference described through vertices A, B, and C (§ 223). Draw radii OA, OB, OC, and OD. In ▲ OAB and OCD, OB = OC. And since, by def., polygon ABCDE is equilateral, ABCD. (?) Again, since, by def., polygon ABCDE is equiangular, Or, < OBC = < OCB. .. ▲ ABC – ▲ OBC = ▲ BCD – Z OCB. ZOBA=≤ OCD. .. Δ ΟΑΒ = 40CD. .. OA= OD. (?) (?) (?) Then, the circumference which passes through A, B, and C also passes through D. In like manner, it may be proved that the circumference which passes through B, C, and D also passes through E. Hence, a O can be circumscribed about ABCDE. Again, since AB, BC, CD, etc., are equal chords of the circumscribed O, they are equally distant from O. (§ 164) Hence, a described with O as a centre, and a line OF to any side AB as a radius, will be inscribed in ABCDE. 341. Def. The centre of a regular polygon is the common centre of the circumscribed and inscribed circles. The angle at the centre is the angle between the radii drawn to the extremities of any side; as AOB. The radius is the radius of the circumscribed circle, OA. The apothem is the radius of the inscribed circle, OF. 342. Cor. From the equal A OAB, OBC, etc., we have ▲ AOB = Z BOC = ≤ COD, etc. But the sum of these is four rt. . (?) (§ 35) Whence, the angle at the centre of a regular polygon is equal to four right angles divided by the number of sides. EXERCISES. Find the angle, and the angle at the centre, 1. Of a regular pentagon. 2. Of a regular dodecagon. 3. Of a regular polygon of 32 sides. 4. Of a regular polygon of 25 sides. PROP. II. THEOREM. 343. If the circumference of a circle be divided into any number of equal arcs, I. Their chords form a regular inscribed polygon. II. Tangents at the points of division form a regular circumscribed polygon. Given circumference ACD divided into five equal arcs, AB, BC, CD, etc., and chords AB, BC, etc. Also, lines LF, FG, etc., tangent to O ACD at A, B, etc., respectively, forming polygon FGHKL. To Prove polygons ABCDE and FGHKL regular. Proof. Chord AB chord BC chord CD, etc. (§ 158) = = Again, arc BCDE: =arc CDEA = arc DEAB, etc., for each is the sum of three of the equal arcs AB, BC, etc. .. / EAB= ≤ ABC = ▲ BCD, etc. Therefore, polygon ABCDE is regular. Again, in ▲ ABF, BCG, CDH, etc., we have AB=BC= CD, etc. (§ 193) (§ 339) Also, since arc AB- arc BC= arc CD, etc., we have BAFZABF=/ CBG=/ BCG, etc. ($ 197) Whence, ABF, BCG, etc., are equal isosceles A. (§§ 68,96) (?) Therefore, polygon FGHKL is regular. 344. Cor. I. 1. If from the middle point of each arc subtended by a side of a regular inscribed polygon lines be drawn to its extremities, a regular inscribed polygon of double the number of sides is formed. 2. If at the middle point of each arc included between two consecutive points of contact of a regular circumscribed polygon tangents be drawn, a regular circumscribed polygon of double the number of sides is formed. 345. Cor. II. An equilateral polygon inscribed in a circle is regular; for its sides subtend equal arcs. (?) 346. Tangents to a circle at the middle points of the arcs subtended by the sides of a regular inscribed polygon, form a regular circumscribed polygon. Given ABCDE a regular polygon inscribed in O AC, and A'B'C' D'E' a polygon whose sides A'B', B'C', etc., are AC at the middle points F, G, etc., of arcs tangent to AB, BC, etc., respectively. To Prove A'B'C'D'E' a regular polygon. 347. Regular polygons of the same number of sides are (The polygons fulfil the conditions of similarity given in § 252.) PROP. V. THEOREM. 348. The perimeters of two regular polygons of the same number of sides are to each other as their radii, or as their apothems. Given P and P' the perimeters, R and R' the radii, and r and r' the apothems, respectively, of regular polygons AC and A'C" of the same number of sides. Proof. Let O be the centre of polygon AC, and O' of A'C', and draw lines OA, OB, O'A', and O'B'. Also, draw line OF 1 AB, and line O'F' A'B'. ZAOB = ZA'O'B'. (§ 342) |